Question

The wavelength of the first line in Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line?

• A. $\frac{20}{27}\lambda$
• B. $\frac{3}{16}\lambda$
• C. $\frac{5}{36}\lambda$
• D. $\frac{3}{4}\lambda$

Answer 1

The correct option is A $\frac{20}{27}\lambda$

The first line in Balmer series corresponds to $n=3$.
$\therefore \frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{3^2})=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}$
$\Rightarrow \lambda =\frac{36}{5R}$
For the second line, $n=4$.
$\therefore \frac{1}{{\lambda }^{\prime }}=R(\frac{1}{2^2}-\frac{1}{4^2})=R(\frac{1}{4}-\frac{1}{16})=\frac{3R}{16}$
$\Rightarrow {\lambda }^{\prime }=\frac{16}{3R}$
Now, $\frac{{\lambda }^{\prime }}{\lambda }=\frac{16}{3R}\times \frac{5R}{36}=\frac{20}{27}\Rightarrow {\lambda }^{\prime }=\frac{20}{27}\lambda$
Hence, the correct choice is (a).